verify that the equations are identities: 1+cosy/1-cosy = sin^2y/(1-cos^2y)
Hint: sin^2y = 1 - cos^2y
I'm going to show you my steps, although, it might not be the way you learned how to do these.
I worked through it and got 1-cos^2y/1-cos^y
I don't believe it is an identity according to my math
this is how I done it 1-cos^2y/(1-cos^2y)(1+cos^2y)= 1-cos^2y/1-cos^2y
I'm telling you that I don't believe the identity is true
I can show you my work. It's pretty simple
How do you decide which side to start on?
I'll show you what I did
I knew that sin^2y = 1 - cos^2y so I substituted that in: \(\large\frac{1+cos x}{1-cos x} = \frac{1-cos^2x}{1-cos^2x} \)
Now look at the right side. It cancels to 1
So you have \(\large\frac{1 + cos x}{1-cos x} = 1\)
Now you multiply both sides by 1 - cos x to get 1 + cos x = 1 - cos x
and now you can see that \(1 + cos x \ne 1 - cos x\)
so the identity is false
thanks
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